Two finite-duration discrete-time signals x[n] and h[n] are defined as: x[n] = 1 for 0 <= n <= 9, and h[n] = 1 for 0 <= n <= N. If y[n] = x[n] * h[n] and y[4]=5 and y[14]=0, find the possible values o

Step 1: Define the convolution sum and the non-zero ranges of the signals. The convolution sum y[n] is given by: y[n] = _k=-^ x[k] h[n-k] The signal x[n] is non-zero for 0 n 9, so x[k]=1 for 0 k 9. The signal h[n] is non-zero for 0 n N, so h[n-k]=1 for 0 n-k N. This implies n-N k n. Step 2: Determine the effective limits of summation for k. For x[k]h[n-k] to be non-zero, k must satisfy both conditions: 0 k 9 n-N k n Combining these, the summation limits for k are: (0, n-N) k (9, n) Since x[k]=1 and h[n-k]=1 within these ranges, y[n] is the number of terms in the sum: y[n] = (9, n) - (0, n-N) + 1 Step 3: Use the condition y[4]=5. Substitute n=4 into the expression for y[n]: y[4] = (9, 4) - (0, 4-N) + 1 y[4] = 4 - (0, 4-N) + 1 y[4] = 5 - (0, 4-N) Given y[4]=5: 5 = 5 - (0, 4-N) (0, 4-N) = 0 For (0, 4-N) to be 0, we must have 4-N 0, which means N 4. We are also given that N 9, so the first constraint is 4 N 9. Step 4: Use the condition y[14]=0. The convolution y[n] is non-zero for n in the range [0+0, 9+N], i.e., 0 n 9+N. If y[14]=0, it means that n=14 is outside the non-zero range of y[n]. Therefore, 14 > 9+N. 14 - 9 > N 5 > N So, N < 5. Step 5: Determine the value of N. We have two constraints for N: 1. 4 N 9 2. N < 5 Combining these, N must be an integer such that 4 N < 5. The only integer that satisfies this condition is N=4. Step 6: Verify the result. If N=4: y[4] = (9, 4) - (0, 4-4) + 1 = 4 - (0, 0) + 1 = 4 - 0 + 1 = 5. (Correct) y[14] = (9, 14) - (0, 14-4) + 1 = 9 - (0, 10) + 1 = 9 - 10 + 1 = 0. (Correct) The value of N is 4.